Bourbaki marks this one as harder than the rest, it is set in small type as supplementary.
Let (Mn), (Pn) be two sequences of mutually disjoint finite sets (not all empty), indexed by the set Z of rational integers. Let αn=Card (Mn), βn=Card (Pn). Suppose that there exists an integer k>0 such that for each n∈Z and each integer l⩾1 we have
αn+αn+1+⋯+αn+l⩽βn−k+βn−k+1+⋯+βn+l+k,
βn+βn+1+⋯+βn+l⩽αn−k+αn−k+1+⋯+αn+l+k.
Let M be the union of the family (Mn) and let P be the union of the family (Pn). Show that there exists a bijection φ of M onto P such that
φ(Mn)⊂i=n−k−1⋃n+k+1Piandφ−1(Pn)⊂i=n−k−1⋃n+k+1Mi
for each n∈Z. (Consider a total ordering on each Mn (resp. Pn) and take M (resp. P) to be the ordinal sum (§ 1, Exercise 3) of the family (Mn)n∈Z (resp. (Pn)n∈Z). If n0 is an index such that Mn0=∅, consider the isomorphisms of M onto P which transform the least element of Mn0 into one of the elements of j=n0−k⋃n0+kPj, and show that one of these isomorphisms satisfies the required conditions. Let δ be the least of the numbers
βn−k+βn−k+1+⋯+βn+l+k−(αn+αn+1+⋯+αn+l),
αn−k+αn−k+1+⋯+αn+l+k−(βn+βn+1+⋯+βn+l)
for all n∈Z and all l⩾1. If n∈Z and l⩾1 are such that, for example, βn−k+βn−k+1+⋯+βn+l+k=δ+αn+αn+1+⋯+αn+l, we may take φ to be such that the least element of Pn−k is the image under φ of the least element of Mn.)
Nothing.